The following reaction occurs in acidic medium
$KMnO_{4} + 8 H^{+} + 5 e^{-} ⟶ K^{+} + Mn^{2 +} + 4 H_{2} O$
What is the equivalent weight of $KMnO_{4}$ ?
(Molecular weight of $KMnO_{4} = 158$ )

Question

The following reaction occurs in acidic medium KMnO_4+8H^++5e^- ⟶K^++ Mn^2++4H_2O What is the equivalent weight of KMnO_4 ? (Molecular weight of KMnO_4=158 )

Accepted Answer

Correct answer: B. (b) KMnO_4+8H^++5e^- ⟶K^++ Mn^2++4H_2O KMnO_4 gains 5 electrons from reductant. ∴ Equivalent weight of KMnO_4 = 1585=31.6 (in g/ equivalent)

$The following reaction occurs in acidic medium KMnO_{4} + 8 H^{+} + 5 e^{-} ⟶ K^{+} + Mn^{2 +} + 4 H_{2} O What is the equivalent weight of KMnO_{4} ? (Molecular weight of KMnO_{4} = 158)$

Solution

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